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Suppose that I have a hollow conductor with no specific format.

I know that if I put a charge inside it, it will induce charge in the inner surface and on the outer surface a charge with a opposite sign.

And that if the conductor is a sphere or a circle, it doesn't matter where inside the charge is, the field outside will stay the same, because of the spherical shell theorem.

My question is, what about a non-spherical conductor? Will the field outside change when I move the charge inside? Why?

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No. The simplest way to see this is that all parts on the outside surface of the conductor will remain at the same potential, irrespective of the location of the charge inside the hollow part.

Thus, if you think of solving first for the potential outside the conductor, and obtaining the field from the gradient of the potential, then the boundary conditions on the outside of the conductor remain unchanged by the motion of the charge inside. Because the boundary conditions are identical, the potentials will be identical and thus the fields must be identical by the uniqueness theorem.

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  • $\begingroup$ I ended up with another argument which seems plausible, the charge induced on the inner surface by the point charge placed inside will by itself cancel any electric field entering the conductor, so the excessive positive charges that are left can distribute themselves as in a positively charged conductor. $\endgroup$
    – Socrates
    Commented Apr 17, 2019 at 20:17
  • $\begingroup$ @HenriqueLengler yes the field will remain $0$ inside but with this argument it's difficult to see how moving a charge elsewhere inside will not affect the charge distribution on the inside and outside of the surfaces of the conductor $\endgroup$
    – ZeroTheHero
    Commented Apr 17, 2019 at 20:26
  • $\begingroup$ What I did not understand about your argument is why "all parts on the outside surface of the conductor will remain at the same potential"? $\endgroup$
    – Socrates
    Commented Apr 17, 2019 at 21:45
  • $\begingroup$ @HenriqueLengler because there is no $\vec E$-field inside a conductor is follows that all parts of a conductor are at the same potential. $\endgroup$
    – ZeroTheHero
    Commented Apr 17, 2019 at 22:01
  • $\begingroup$ But if the charge inside change and the distribution of positive charges on the surface change, the boundary condition will change, won't it? $\endgroup$
    – Socrates
    Commented Apr 17, 2019 at 22:45

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